-6x+7x(1-x)=4(x-4)

Solution for -6x+7x(1-x)=4(x-4) equation:

-6x+7x(1-x)=4(x-4)

We move all terms to the left:

-6x+7x(1-x)-(4(x-4))=0

We add all the numbers together, and all the variables

-6x+7x(-1x+1)-(4(x-4))=0

We multiply parentheses

-7x^2-6x+7x-(4(x-4))=0


We calculate terms in parentheses: -(4(x-4)), so:

4(x-4)

We multiply parentheses

4x-16

Back to the equation:

-(4x-16)


We add all the numbers together, and all the variables

-7x^2+x-(4x-16)=0

We get rid of parentheses

-7x^2+x-4x+16=0

We add all the numbers together, and all the variables

-7x^2-3x+16=0

a = -7; b = -3; c = +16;
Δ = b2-4ac
Δ = -32-4·(-7)·16
Δ = 457
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{457}}{2*-7}=\frac{3-\sqrt{457}}{-14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{457}}{2*-7}=\frac{3+\sqrt{457}}{-14}
$